Serializability and strict schedules

Question

Answer is given as D. Not serializable, agreed. But how is it strict??

Comments

It is not a strict schedule because definition of strict schedule saying a value written by one transaction t1 can not be read or over written by another transactions until the transaction t1 is commited or aborted.But here transaction t1 is overwriting the value of X that is written by Transaction t2 before t2 commits.

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@akash.dinkar12 

Let's say in this schedule 

R1(X), R2(X), W1(X), C2, C1 

according to you this is also not strict schedule . but i think this is strict schedule. correct me if i am wrong

my definition for strict schedule :- Suppose if in a schedule(S) transaction T1 perform W(X) then T2 is not allowed to R(X) or W(X) till T1 commits or rollback . 

it means in strict schedule you will not find any Write-Read , Write-Write conflict .

@Arjun Sir please see this is correct ?

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ur definition of strict schedule is correct .

However we can clearly see a write – write conflict between write of t(2) to write of t1.

 

or in other words first write is performed by t2 but before it can commit , t1 also performs a write.

Answer 1

Option (B)

If a precedence graph is drawn for above schedule, we see there is an edge T1 --> T2 and also T2 --> T1. This makes a cycle. Thus, it is not conflict serializable.

It is not strict as T1 is overriding X written by T2 before letting T1 commit and hence violating the definition of strict schedule.

Comments

To check whether a schedule is Serializable or not you have to check whether a schedule is view serializable or not , because every view serializable schedule is Serializable and vice-versa . 

@Pratik Gawali your reason for 1st part is correct but in the second part " It is not strict as T1 is overriding X written by T2 before letting T2 commit and hence violating the definition of a strict schedule."

This should be correct.